POJ2299归并排序

C - Ultra-QuickSort（7.2.2）（7.2应用排序算法编程的实验范例）

Crawling in process...Crawling failedTime Limit:7000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230

Sample Output

6
0
#include <cstdio>#include <cstring>#include <iostream>#define lolo long longconst lolo maxn=510000;using namespace std;lolo ans;int a[maxn],t[maxn],n;void _sort(int l,int r)//归并排序{    if(l==r)return;    int mid=(l+r)/2;    _sort(l,mid);    _sort(mid+1,r);    int i=l,j=mid+1,now=0;    while(i<=mid&&j<=r)    {        if(a[i]>a[j])        {            ans+=mid-i+1;            t[++now]=a[j++];        }        else        {            t[++now]=a[i++];        }    }    while (i<=mid)t[++now]=a[i++];    while (j<=r)t[++now]=a[j++];    now=0;    for(int k=l; k<=r; k++)a[k]=t[++now];}int main(){    scanf("%d",&n);    while(n)    {        for(int i=1; i<=n; ++i)scanf("%d",&a[i]);        ans=0;        _sort(1,n);        cout<<ans<<'\12';        scanf("%d",&n);    }    return 0;}

学习总结：归并排序